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3 years ago

13

Fusion and fission are both processes that involve which fundamental force? to ask? A. gravitational B. weak nuclear C. electrom

agnetic D. strong nuclear

1 answer:

marin [14]3 years ago

5 0

Fusion & Fission are both processes that involve D. Strong Nuclear Fundamental forces
"D". is the answer

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• List four uses of the concave mirror.<br> State five uses of lenses in everyday activities

Zina [86]

FOUR USES OF CONCAVE MIRROR:Satellite dishes,headlights of a car, telescopes used for astronomical studies, and shaving mirrors because of there curved and reflective surface.

FIVE USES OF LENSES: Camera lens ,microscopes ,magnifying glass,eyeglasses,projector

6 0

2 years ago

Elements that typically give up electrons CHECK ALL THAT APPLY

lys-0071 [83]

Answer:

B. have a lower ionization energy

D. are metals

Explanation:

An atom can be defined as the smallest unit comprising of matter that forms all chemical elements. Thus, atoms are basically the building blocks of matters and as such determines or defines the structure of a chemical element.

Generally, atoms are typically made up of three distinct particles and these are protons, neutrons and electrons.

In Chemistry, electrons can be defined as subatomic particles that are negatively charged and as such has a magnitude of -1.

Valence electrons can be defined as the number of electrons present in the outermost shell of an atom. Valence electrons are used to determine whether an atom or group of elements found in a periodic table can bond with others. Thus, this property is typically used to determine the chemical properties of elements.

Valency can be defined as a measure of the combining power of a chemical element with other atoms to form a molecule or chemical compound.

Typically, valency is measured by the amount of hydrogen atoms that a chemical element can combine with or displace to form a molecule or chemical compound.

Ionization energy can be defined as the minimum energy required to remove or detach an electron from a neutral atom in a gaseous state.

Generally, the ionization energy of chemical elements tend to increase from left to right across a period on the periodic table. This increase is due to the fact that the atomic radius of chemical elements generally decreases across the periodic table, typically from alkali metals (group one elements such as hydrogen, lithium and sodium) to noble gases (group eight elements such as argon, helium and neon) i.e from left to the right of the periodic table. Also, the atomic radius of a chemical element increases down each group of the periodic table, typically from top to bottom (column).

This ultimately implies that, atoms with relatively large atomic radii tend to have a low electron affinity and a low ionization energy.

In conclusion, chemical elements that typically give up electrons are metals because their outermost shell contains excess electrons and have a lower ionization energy.

4 0

2 years ago

Mixing chemicals in a cold pack creates a reaction that:

MatroZZZ [7]

Honestly i don’t know but I’d just guess if I were you

7 0

3 years ago

Read 2 more answers

1. What was the Michelson-Morley experiment designed to do?2. When was the Michelson-Morley experiment done?3. What was the ethe

babymother [125]

Answer:

1) designed to measure the difference in speed of light in different directions , 1887

Explanation:

1) This experiment was designed to measure the difference in speed of light in different directions and therefore find the speed of the ether.

2) was made in 1887

3) At that time it was assumed that it was the medium in which light traveled and it is everywhere

4) the speed of the wave depends on the characteristics of the medium where it travels,

for the one in a string depends on the tension and density

for an electromagnetic wave of the permittivity and permeability of the vacuum

5) In this type of interferometer the beam is divided into two rays

6) In his interrupter, he had to accurately measure the displacement of the fringes in a telescope, for which he had to minimize vibrations, he had problems in the movement of one of the arms, changes in temperature

7) In Michelsom's second experiment, the apparatus could measure 0.01 fringes by increasing the length of the arms by 11 m

8) The new interferometer floated on a bed of mercury

9) Couldn't measure any difference in speed of light in different directions

10) Physics was forced to eliminate the concept of ETHER

11) One of the principles of relativities that the speed of light is constant in all inertial efficiency systems

12) Michelson in 1907

13) It seems that Einstein did not know the results of this experiment

6 0

3 years ago

Illustrates an Atwood's machine. Let the masses of blocks A and B be 7.00 kg and 3.00 kg , respectively, the moment of inertia o

Harman [31]

Answer:

A) 1.55

B) 1.55

C) 12.92

D) 34.08

E) 57.82

Explanation:

The free body diagram attached, R is the radius of the wheel

Block B is lighter than block A so block A will move upward while A downward with the same acceleration. Since no snipping will occur, the wheel rotates in clockwise direction.

At the centre of the whee, torque due to B is given by

${\tau _2} = - {T_{\rm{B}}}R$

Similarly, torque due to A is given by

${\tau _1} = {T_{\rm{A}}}R$

The sum of torque at the pivot is given by

$\tau = {\tau _1} + {\tau _2}$

Replacing ${\tau _1}$ and ${\tau _2}$ by ${T_{\rm{A}}}R$ and $- {T_{\rm{B}}}R$ respectively yields

$\begin{array}{c}\\\tau = {T_{\rm{A}}}R - {T_{\rm{B}}}R\\\\ = \left( {{T_{\rm{A}}} - {T_{\rm{B}}}} \right)R\\\end{array}$

Substituting $I\alpha$ for $\tau$ in the equation $\tau = \left( {{T_{\rm{A}}} - {T_{\rm{B}}}} \right)R$

$I\alpha=\left( {{T_{\rm{A}}} - {T_{\rm{B}}}} \right)R$

$\frac{I\alpha}{R} =\left {{T_{\rm{A}}} - {T_{\rm{B}}}} \right$

The angular acceleration of the wheel is given by $\alpha = \frac{a}{R}$

where a is the linear acceleration

Substituting $\frac{a}{R}$ for $\alpha$ into equation

$\frac{I\alpha}{R} =\left {{T_{\rm{A}}} - {T_{\rm{B}}}} \right$ we obtain

$\frac{Ia}{R^2} =\left {{T_{\rm{A}}} - {T_{\rm{B}}}} \right$

Net force on block A is

${F_{\rm{A}}} = {m_{\rm{A}}}g - {T_{\rm{A}}}$

Net force on block B is

${F_{\rm{B}}} = {T_{\rm{B}}} - {m_{\rm{B}}}g$

Where g is acceleration due to gravity

Substituting ${m_{\rm{B}}}a$ and ${m_{\rm{A}}}a$ for ${F_{\rm{B}}}$ and ${F_{\rm{A}}}$ respectively into equation $\frac{Ia}{R^2} =\left {{T_{\rm{A}}} - {T_{\rm{B}}}} \right$ and making a the subject we obtain

$\begin{array}{c}\\{m_{\rm{A}}}g - {m_{\rm{A}}}a - \left( {{m_{\rm{B}}}g + {m_{\rm{B}}}a} \right) = \frac{{Ia}}{{{R^2}}}\\\\\left( {{m_{\rm{A}}} - {m_{\rm{B}}}} \right)g - \left( {{m_{\rm{A}}} + {m_{\rm{B}}}} \right)a = \frac{{Ia}}{{{R^2}}}\\\\\left( {{m_{\rm{A}}} + {m_{\rm{B}}} + \frac{I}{{{R^2}}}} \right)a = \left( {{m_{\rm{A}}} - {m_{\rm{B}}}} \right)g\\\\a = \frac{{\left( {{m_{\rm{A}}} - {m_{\rm{B}}}} \right)g}}{{\left( {{m_{\rm{A}}} + {m_{\rm{B}}} + \frac{I}{{{R^2}}}} \right)}}\\\end{array}$

Since ${m_{\rm{B}}}$ = 3kg and ${m_{\rm{B}}}$ = 7kg

g=9.81 and R=0.12m, I=0.22 ${\rm{ kg}} \cdot {{\rm{m}}^2}$

Substituting these we obtain

$a = \frac{{\left( {{m_{\rm{A}}} - {m_{\rm{B}}}} \right)g}}{{\left( {{m_{\rm{A}}} + {m_{\rm{B}}} + \frac{I}{{{R^2}}}} \right)}}$

$\begin{array}{c}\\a = \frac{{\left( {7{\rm{ kg}} - 3{\rm{ kg}}} \right)\left( {9.81{\rm{ m/}}{{\rm{s}}^2}} \right)}}{{\left( {7{\rm{ kg}} + 3{\rm{ kg}} + \frac{{0.22{\rm{ kg/}}{{\rm{m}}^2}}}{{{{\left( {0.120{\rm{ m}}} \right)}^2}}}} \right)}}\\\\ = 1.55235{\rm{ m/}}{{\rm{s}}^2}\\\end{array}$

Therefore, the linear acceleration of block A is 1.55 ${\rm{ m/}}{{\rm{s}}^2}$

(B)

For block B

${a_{\rm{B}}} = {a_{\rm{A}}}$

Therefore, the acceleration of both blocks A and B are same

1.55 ${\rm{ m/}}{{\rm{s}}^2}$

(C)

The angular acceleration is $\alpha = \frac{a}{R}$

$\begin{array}{c}\\\alpha = \frac{{1.55{\rm{ m/}}{{\rm{s}}^2}}}{{0.120{\rm{ m}}}}\\\\ = 12.92{\rm{ rad/}}{{\rm{s}}^2}\\\end{array}$

(D)

Tension on left side of cord is calculated using

$\begin{array}{c}\\{T_{\rm{B}}} = {m_{\rm{B}}}g + {m_{\rm{B}}}a\\\\ = {m_{\rm{B}}}\left( {g + a} \right)\\\end{array}$

$\begin{array}{c}\\{T_{\rm{B}}} = \left( {3{\rm{ kg}}} \right)\left( {9.81{\rm{ m/}}{{\rm{s}}^2} + 1.55{\rm{ m/}}{{\rm{s}}^2}} \right)\\\\ = 34.08{\rm{ N}}\\\end{array}$

(E)

Tension on right side of cord is calculated using

$\begin{array}{c}\\{T_{\rm{A}}} = {m_{\rm{A}}}g - {m_{\rm{A}}}a\\\\ = {m_{\rm{A}}}\left( {g - a} \right)\\\end{array}$

$\begin{array}{c}\\{T_{\rm{A}}} = \left( {7{\rm{ kg}}} \right)\left( {9.81{\rm{ m/}}{{\rm{s}}^2} – 1.55{\rm{ m/}}{{\rm{s}}^2}} \right)\\\\ = 57.82{\rm{ N}}\\\end{array}$

6 0

2 years ago